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Diketahui Ksp​ X(OH)2​ dalam air adalah 3,2×10−11. Larutan jenuh  akan mengendap pada pH sebesar ....

Pertanyaan

Diketahui begin mathsize 14px style K subscript sp end style begin mathsize 14px style X open parentheses O H close parentheses subscript 2 end style dalam air adalah begin mathsize 14px style 3 comma 2 cross times 10 to the power of negative sign 11 end exponent end style. Larutan jenuh begin mathsize 14px style X open parentheses O H close parentheses subscript 2 end style akan mengendap pada pH sebesar ....space 

  1. 12 + log 4space 

  2. 10 + log 4space 

  3. 9 + log 2space 

  4. 8 + log 4space 

  5. 7 + log 2space 

Pembahasan Soal:

Larutan akan mengendap apabila nilai begin mathsize 14px style Q subscript sp end style>begin mathsize 14px style K subscript sp end style. Nilai begin mathsize 14px style K subscript sp end style begin mathsize 14px style X open parentheses O H close parentheses subscript 2 end style = begin mathsize 14px style 3 comma 2 cross times 10 to the power of negative sign 11 end exponent end style

Langkah pertama yaitu menentukan kelarutan (s) dari begin mathsize 14px style K subscript sp end style
 

begin mathsize 14px style X open parentheses O H close parentheses subscript 2 equilibrium X to the power of 2 plus sign and 2 O H to the power of minus sign space space s space space space space space space space space space space space space space s space space space space space space space space space 2 s end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp end cell equals cell open square brackets X to the power of 2 plus sign close square brackets open square brackets O H to the power of minus sign close square brackets squared end cell row cell K subscript sp end cell equals cell s middle dot left parenthesis 2 s right parenthesis squared K subscript sp end subscript equals 4 s cubed end cell row s equals cell cube root of K subscript sp over 4 end root end cell row s equals cell cube root of fraction numerator 3 comma 2 cross times 10 to the power of negative sign 11 end exponent over denominator 4 end fraction end root end cell row s equals cell cube root of 0 comma 8 cross times 10 to the power of negative sign 11 end exponent end root end cell row s equals cell cube root of 8 cross times 10 to the power of negative sign 12 end exponent end root end cell row s equals cell 2 cross times 10 to the power of negative sign 4 end exponent end cell end table end style 
 

Langkah selanjutnya adalah menentukan pH sebagai berikut.
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 2 s end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 2 open parentheses 2 cross times 10 to the power of negative sign 4 end exponent close parentheses end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 4 cross times 10 to the power of negative sign 4 end exponent end cell row blank blank blank row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row pOH equals cell negative sign log space 4 cross times 10 to the power of negative sign 4 end exponent end cell row pOH equals cell 4 minus sign log space 4 end cell row blank blank blank row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign left parenthesis 4 minus sign log space 4 right parenthesis end cell row pH equals cell 10 plus log space 4 end cell end table end style 
 

Jadi, jawaban yang benar adalah B.space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Ibabas

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Diketahui Ksp​ Cr(OH)3​ dalam air adalah . Larutan jenuh  akan mengendap pada pH sebesar ....

Pembahasan Soal:

Persamaan begin mathsize 14px style K subscript sp end style begin mathsize 14px style Cr open parentheses O H close parentheses subscript 3 end style adalah sebagai berikut.
 

Cr open parentheses O H close parentheses subscript 3 left parenthesis italic s right parenthesis rightwards harpoon over leftwards harpoon Cr to the power of 3 plus sign left parenthesis italic a italic q right parenthesis plus 3 O H to the power of minus sign left parenthesis italic a italic q right parenthesis space space space space space space space space space space space space space space space space space space space space space space space s space space space space space space space space space space space space space space space space space 3 s  
 

Kelarutan (s) begin mathsize 14px style Cr open parentheses O H close parentheses subscript 3 end style dapat dihitung menggunakan persamaan berikut.
 

  Error converting from MathML to accessible text. 


Dengan demikian, larutan jenuh begin mathsize 14px style Cr open parentheses O H close parentheses subscript 3 end style akan mengendap pada pH yang dapat dihitung sebagai berikut.
 

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 3 s end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 3 cross times 10 to the power of negative sign 2 end exponent end cell row blank blank blank row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row pOH equals cell negative sign log space left parenthesis 3 cross times 10 to the power of negative sign 2 end exponent right parenthesis end cell row pOH equals cell 2 minus sign log space 3 end cell row blank blank blank row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign left parenthesis 2 minus sign log space 3 right parenthesis end cell row pH equals cell 12 plus log space 3 end cell end table    
 

Jadi, jawaban yang benar adalah A.

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Kelarutan suatu garam Cu(OH)2​ dalam air adalah 8×10−4mol/L. pH larutan jenuh  adalah ….

Pembahasan Soal:

Error converting from MathML to accessible text. 

Error converting from MathML to accessible text. 

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Diketahui Ksp L(OH)3​ dalam air adalah . Larutan jenuh  akan mengendap pada pH sebesar ….

Pembahasan Soal:

Error converting from MathML to accessible text.

Error converting from MathML to accessible text.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets to the power of blank end cell equals cell 3 s equals 3 cross times 10 to the power of negative sign 3 end exponent end cell row pOH equals cell 3 minus sign log invisible function application 3 end cell row pH equals cell 14 minus sign open parentheses 3 minus sign log 3 close parentheses equals 11 plus log 3 end cell end table end style 

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Roboguru

Kelarutan garam L(OH)2​ dalam air adalah 5×10−4mol/L. pH larutan jenuh  adalah ....

Pembahasan Soal:

Reaksi kesetimbangan garam begin mathsize 14px style L open parentheses O H close parentheses subscript 2 end style dalam air adalah sebagai berikut.


L open parentheses O H close parentheses subscript 2 left parenthesis italic s right parenthesis equilibrium L to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 O H to the power of minus sign left parenthesis italic a italic q right parenthesis 5 cross times 10 to the power of negative sign 4 end exponent space space space space space space space space space space space space space space space space space space space space space space space 2 cross times 5 cross times 10 to the power of negative sign 4 end exponent
 


Nilai konsetrasi ion O H to the power of minus sign sudah diketahui sehingga pH larutan dapat dihitung sebagai berikut.


table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 2 cross times 5 cross times 10 to the power of negative sign 3 end exponent end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 10 to the power of 1 cross times 10 to the power of negative sign 4 end exponent end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 10 to the power of negative sign 3 end exponent end cell row blank blank blank row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row pOH equals cell negative sign log space open square brackets 10 to the power of negative sign 3 end exponent close square brackets end cell row pOH equals 3 row blank blank blank row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign 3 end cell row pH equals 11 end table


Jadi, jawaban yang tepat adalah A.

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Pada suhu 25∘C diketahui hasil kali kelarutan dari perak nitrat (AgNO3​) dalam air adalah 1,8×10−15. Kelarutan garam dalam gram/L adalah .... (Ar​ Ag = 108,5; N = 14; O = 16)

Pembahasan Soal:

Langkah pertama yang dilakukan adalah menghitung kelarutan (s) Ag N O subscript 3 dalam mol/L dengan menggunakan persamaan berikut.
 

Ag N O subscript 3 equilibrium Ag to the power of plus sign and N O subscript 3 to the power of minus sign space space space space s space space space space space space space space space space space s space space space space space space space space space s 

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ag N O subscript 3 end cell equals cell open square brackets Ag to the power of plus sign close square brackets open square brackets N O subscript 3 to the power of minus sign close square brackets end cell row cell K subscript sp space Ag N O subscript 3 end cell equals cell open parentheses s close parentheses open parentheses s close parentheses end cell row cell K subscript sp thin space Ag N O subscript 3 end cell equals cell s squared end cell row s equals cell square root of K subscript sp space Ag N O subscript 3 end root end cell row s equals cell square root of 1 comma 8 cross times 10 to the power of negative sign 15 end exponent end root end cell row s equals cell square root of 18 cross times 10 to the power of negative sign 16 end exponent end root end cell row s equals cell 4 comma 2 cross times 10 to the power of negative sign 8 end exponent space mol forward slash L end cell end table  
 

Langkah selanjutnya adalah mengubah kelarutan (s) dalam mol/L menjadi g/L dengan menggunakan persamaan berikut.  
 

table attributes columnalign right center left columnspacing 0px end attributes row cell s space Ag N O subscript 3 space left parenthesis g forward slash L right parenthesis end cell equals cell s space Ag N O subscript 3 space left parenthesis mol forward slash L right parenthesis cross times italic M subscript r space Ag N O subscript 3 space left parenthesis g forward slash mol right parenthesis end cell row cell s space Ag N O subscript 3 space left parenthesis g forward slash L right parenthesis end cell equals cell 4 comma 2 cross times 10 to the power of negative sign 8 end exponent space fraction numerator up diagonal strike mol over denominator L end fraction cross times 170 space fraction numerator g over denominator up diagonal strike mol end fraction end cell row cell s space Ag N O subscript 3 space left parenthesis g forward slash L right parenthesis end cell equals cell 714 cross times 10 to the power of negative sign 8 end exponent space g forward slash L end cell row cell s space Ag N O subscript 3 space left parenthesis g forward slash L right parenthesis end cell equals cell 7 comma 1 cross times 10 to the power of negative sign 6 end exponent space g forward slash L end cell end table 
 

Jadi, jawaban yang tepat adalah C.space  

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